cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A265742 Numbers k such that there exists at least an integer in the interval [e*k - 1/k, e*k + 1/k] where e = 2.71828... is Euler's number.

Original entry on oeis.org

1, 2, 3, 4, 7, 14, 32, 39, 71, 142, 465, 536, 1001, 2002, 3003, 8544, 9545, 18089, 36178, 54267, 190435, 208524, 398959, 797918, 1196877, 4996032, 5394991, 10391023, 20782046, 31173069, 150869313, 161260336, 312129649, 624259298, 936388947, 1248518596, 5155334720
Offset: 1

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Author

Michel Lagneau, Dec 15 2015

Keywords

Comments

Conjecture: the sequence is infinite.
See the reference for a similar problem with Fibonacci numbers.
The corresponding integers in each interval [e*k - 1/k, e*k + 1/k] are 2, 3, 5, 8, 11, 19, 38, 87, 106, 193, 386, 1264, ... (see A265741).
For k > 1, the interval [e*k - 1/k, e*k + 1/k] contains exactly one integer.
We observe two properties:
(1) a(n) = m*a(n-m+1) for some n, m=2,3,4.
Examples:
m = 2 => a(2)=2*a(1), a(6)=2*a(5), a(10)=2*a(9), a(14)=2*a(13), a(19)=2*a(18), a(24)=2*a(23), ...
m = 3 => a(3)=3*a(1), a(15)=3*a(13), a(20)=3*a(18), a(25)=3*a(23), ...
m = 4 => a(4)=4*a(1), a(36)=4*a(33), ...
(ii) a(n+2) = a(n) + a(n+1) for n = 1, 3, 7, 11, 13, 16, 18, 21, 23, 26, 28, ...

Examples

			a(1) = 1 because there exist two integers (2 and 3) in the interval [1*e - 1/1, 1*e + 1/1] = [1.71828..., 3.71828...];
a(2) = 2 because the number 5 belongs to the interval [2*e - 1/2, 2*e + 1/2] = [4.93656..., 5.93656...];
a(3) = 3 because the number 8 belongs to the interval [3*e - 1/3, 3*e + 1/3] = [7.82151..., 8.48817...].
		

Crossrefs

Programs

  • Maple
    *** the program gives the interval [a,b],the integer(s) between [a,b] and k ***
    nn:=10^7:
    e:=exp(1):
    for n from 1 to nn do:
    x1:=evalhf(e*n-1/n):y1:=evalhf(e*n+1/n):
    x:=floor(x1):y:=floor(y1):
    for j from x+1 to y do:
    printf("%g %g %g %g\n",x1,y1,j,n):
    od:
    od:

Extensions

a(31)-a(32) corrected by and more terms from Jinyuan Wang, Jul 16 2025
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